Solution Manual Fundamentals of Physics 11th Edition by Halliday and Resnick and Walker

1. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 Chapter 1 1. THINKIn this problem we’re given the radius of Earth, and asked to compute its circumference, surface area and volume. EXPRESS Assuming Earth to be a sphere of radius  6.37 10 m 10 km m E R   the corresponding circumference, surface area and volume are:      6 3 3 6.37 10 km,  4      2 E 3 E . 2 , 4 , C R A R V R E 3 The geometric formulas are given in Appendix E. ANALYZE (a) Using the formulas given above, we find the circumference to be 2 2 (6.37 E C R     (b) Similarly, the surface area of Earth is  4 4 6.37 E A R      (c) and its volume is  6.37 3 3 LEARN From the formulas given, we see that of volume to surface area, and surface area to circumference are / 2 E A C R  . 2. The conversion factors are: 1 gry 1/10 line  inch. The factors imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2).   4.00 10 km.  3 4 10 km)  2   2 3 8 2 , 10 km 5.10 10 km   4 4  3      3 E 3 12 3 10 km 1.08 10 km . V R 2 E 3 E R , R , and R . The ratios /3 E V A R  C A V E and / , 1 line 1/12 inch  and 1 point = 1/72 2 2 0.50 gry = 0.18 point . 1 complete document is available on https://solumanu.com/ *** contact me if site not loaded

2. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 smtb98@gmail.com smtb98@gmail.com 2 CHAPTER 1 (a) Since 1 km = 1  103 m and 1 m = 1  106m, 1km The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0  109m. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m,  1cm = 10 m = 10 m 10 We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  104. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,  1.0yd = 0.91m 10  4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain 1inch 0.80 cm = 0.80 cm 2.54 cm  (b) With 12 points = 1 pica, we have 1inch 0.80 cm = 0.80 cm 2.54 cm   5. THINK This problem deals with conversion of furlongs to rods and chains, all of which are units for distance. EXPRESS Given that 1 furlong 201.168 m,  the relevant conversion factors are      m m    3 3 6 9 10 m 10 m 10 10 m.     m m    2 2 6 4 10 m.    9.1 10   6 5 m m m.         6 picas 1inch   1.9 picas.           6 picas 1inch 12 points 1 pica   23 points.   and 1chain , 1rod 5.0292 m 20.117 m 1 rod 5.0292 m    1.0 furlong 201.168 m (201.168 m) 40 rods, and 1 chain 20.117 m    . 1.0 furlong 201.168 m (201.168 m) 10 chains Note the cancellation of m (meters), the unwanted unit. ANALYZE Using the above conversion factors, we find 40 rods 1 furlong  d    (a) the distance d in rods to be 4.0 furlongs 4.0 furlongs 160 rods, complete document is available on https://solumanu.com/ *** contact me if site not loaded

3. 3 10 chains 1 furlong  d    (b) and in chains to be 4.0 furlongs 4.0 furlongs 40 chains. LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160 rods (1 rod 5 m  ) and 40 chains (1 chain  6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1 fanega = 1 already completed part) implies that 1 cuartilla = 1 Continuing in this way, the remaining entries in the first column are 6.94  103 and 3 3.47 10  . (b) In the second (“fanega”) column, we find 0.250, 8.33  102, and 4.17  102 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get1 (e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94  103 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86  102 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m3 or 55501 cm3. Thus, 7.00 almudes = 7.00 7. We use the conversion factors found in Appendix D. 1 acre ft = (43,560 ft ) ft = 43,560 ft  Since 2 in. = (1/6) ft, the volume of water that fell during the storm is (26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) V   Thus,   43560 10 . ft acre ft ). So our results make sense. 20 m 12 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the 48 cahiz, or 2.08  102 cahiz. 2 = 0.500 for the last entry. 12 fanega = 7.00 12 (55501 cm3) = 3.24  104 cm3.  2 3  4.66 10 ft .  2 2 2 7 3 7 3 466 . 10 4 ft V     3 11 . 10 acre ft.  3 complete document is available on https://solumanu.com/ *** contact me if site not loaded

4. smtb98@gmail.com 4 CHAPTER 1 smtb98@gmail.com 8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have     (b) In units of Z, we have     9. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = r2/2, where r is the radius. Therefore, the volume is      258 W 212 S    50.0 S 50.0 S 60.8 W    156 Z 180 S    50.0 S 50.0 S 43.3 Z 2 V r z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have 3 10 m 2000km 1km             2 10 cm 1m      5 2000 10 cm. r In these units, the thickness becomes       2 10 cm 1m       2 3000m 3000m 3000 10 cm z      2      5 2 22 3 which yields 2000 10 cm 3000 10 cm 1.9 10 cm . V 2 10. Since a change of longitude equal to 360corresponds to a 24 hour change, then one expects to change longitude by360 /24 15    before resetting one's watch by 1.0 h. 11. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43. (b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864. 12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so complete document is available on https://solumanu.com/ *** contact me if site not loaded

5. 5 b gc h  6 37 . day m 10 86400 gb m m s day   31 . m s . 14 b g 13. The time on any of these clocks is a straight-line function of that on another, with slopes  1 and y-intercepts  0. From the data in the figure we deduce 2 594 , 7 7 These are used in obtaining the following results. (a) We find  40 when t'AtA = 600 s. 2 7 7 b g b g (c) Clock B reads tB = (33/40)(400) (662/5)  198 s when clock A reads tA = 400 s. (d) From tC = 15 = (2/7)tB + (594/7), we get tB245 s. 14. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the inside front cover of the textbook (also Table 1–2).  1 century 10 century 1 century    (b) The percent difference is therefore 52.6 min 50 min 52.6 min 15. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, this is roughly 1.21  1012s. 16. We denote the pulsar rotation rate f (for frequency). 1 rotation 1.55780644887275 33 40 662 5     . t t t t C B B A 33  B  A      495 s t t t t B A 2 (b) We obtain      B   s. 495 141 t t t t C C B                      100 y 365 day 1 y 24 h 1 day 60 min 1 h      6 52.6 min. (a)   4.9%.  f   3 10 s complete document is available on https://solumanu.com/ *** contact me if site not loaded

6. smtb98@gmail.com 6 CHAPTER 1 smtb98@gmail.com (a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations: 1 rotation 1.55780644887275 10   which should now be rounded to 3.88  108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or 1 rotation 1 10 1.55780644887275  which yields the result t = 1557.80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits). (c) Careful reading of the problem shows that the time-uncertainty per revolution is 17 3 10 s   . We therefore expect that as a result of one million revolutions, the uncertainty should be ( 3 10 )(1 10 )= 3 10    17. THINK In this problem we are asked to rank 5 clocks, based on their performance as timekeepers. EXPRESS We first note that none of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important here is that the clock advance by the same (or nearly the same) amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. ANALYZE The chart below gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning. Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from 5 s to +10 s, for clock E it is in the range from 70 s to 2 s. After C and D, A has          604800 s 388238218.4 N  3 s       6 t   3 10 s  17     6 11 . s

7. 7 the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E. CLOCK Sun. Mon. Tues. -Mon. -Tues. -Wed. A 16 16 15 B 3 +5 10 C 58 58 58 D +67 +67 +67 E +70 +55 +2 LEARN Of the five clocks, the readings in clocks A, B and E jump around from one 24- h period to another, making it difficult to correct them. 18. The last day of the 20 centuries is longer than the first day by    20 century 0.001s century The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is  average increase in length of a day number of days T               or roughly two hours. 19. When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B. Let d be the distance from point B to your eyes. From the Pythagorean theorem, we have ( ) d r r h    Wed. -Thurs. 17 +5 58 +67 +20 Thurs. -Fri. 15 +6 58 +67 +10 Fri. -Sat. 15 7 58 +67 +10   0.02 s.   0.01s day 7305 s 365.25 day y   2000 y   rh h  2 2 2 2 2 2 r

8. smtb98@gmail.com 8 CHAPTER 1 smtb98@gmail.com  rh h  2 2 or dropped, leading to points A and B is , which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s. The value of  can be obtained by using    This yields (360 )(11.1 s) (24 h)(60 min/h)(60 s/min) Using tan d r   , we have tan 2 d r    where r is the radius of the Earth. Since r 2 d rh  . Now the angle between the two radii to the two tangent h, the second term can be 2 , d 2 t . 360 24 h      0.04625 . 2 2 2 , or rh 2 h  r  2 tan Using the above value for  and h = 1.7 m, we have 20. (a) We find the volume in cubic centimeters r  5.2 10 m.  6 3            3 231in 1gal 2.54cm 1in   7.31 10 cm  5 3 193gal = 193gal  and subtract this from 1  106 cm3 to obtain 2.69  105 cm3. The conversion gal  in3 is given in Appendix D (immediately below the table of Volume conversions). (b) The volume found in part (a) is converted (by dividing by (100 cm/m)3) to 0.731 m3, which corresponds to a mass of 1000 kg m 0.731 m c hc using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731kg 4.06 0.0018 kg min after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h). 21. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661  1027 kg). Thus, h 3 2 = 731 kg   5 10 min = 0.77 y

9. 9  24 598 1661 . gc . 10 kg M m     49 E 90 . 10 . N b h   27 40 u 10 kg u 22. The density of gold is 19.32g 1 cm m V     3 19.32g/cm . 3 (a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density  = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be m    We convert the volume to SI units:  1.430 cm 100 cm  Since V = Az with z = 1  10-6 m (metric prefixes can be found in Table 1–2), we obtain   1 10 m (b) The volume of a cylinder of length  is V a circle: A = r2. Therefore, with r = 2.500  106 m and V = 1.430  106 m3, we obtain V r  23. THINK This problem consists of two parts: in the first part, we are asked to find the mass of water, given its volume and density; the second part deals with the mass flow rate of water, which is expressed as kg/s in SI units. EXPRESS From the definition of density:   as m V   , the product of the volume of water and its density. With 1 g = 1  103 kg and 1 cm3 = (1  102m)3 = 1  106m3, the density of water in SI units (kg/m3) is 3 3 3 1 g/cm cm g   To obtain the flow rate, we simply divide the total mass of the water by the time taken to drain it. ANALYZE (a) Using m V   , the mass of a cubic meter of water is cm3 1430 . . V 3      1m      3 6 3 1.430 10 m . V  6 3 1430 . 10 m A   2 1430 . m .  6   where the cross-section area is that of A     4 7.284 10 m 72.84 km. 2 , we see that mass can be calculated / m V             3     1 g 10 kg cm     1 10 kg m .   3 3 6 3 10 m

10. smtb98@gmail.com 10 CHAPTER 1 smtb98@gmail.com    (1 10 kg/m )(1m ) 1000kg.   3 3 3 m V (b) The total mass of water in the container is M   and the time elapsed is t = (10 h)(3600 s/h) = 3.6  104 s. Thus, the mass flow rate R is 5.70 3.6 t  LEARN In terms of volume, the drain rate can be expressed as 3 5700 m 3.6 10 s t  The greater the flow rate, the less time required to drain a given amount of water. 24. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2). The surface area A of each grain of sand of radius r = 50 m = 50  106 m is given by A = 4(50  106)2 = 3.14  108 m2 (Appendix E contains a variety of geometry formulas). We introduce the notion of density, / m V   , so that the mass can be found from m = V, where  = 2600 kg/m3. Thus, using V = 4r3/3, the mass of each grain is 4 4 kg 2600 3 m     We observe that (because a cube has six equal faces) the indicated surface area is 6 m2. The number of spheres (the grains of sand) N that have a total surface area of 6 m2 is given by 2 6 m 3.14 10  Therefore, the total mass M is  1.91 10 M Nm   25. The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0106 m3. Letting “d” stand for the thickness of the mud after it has (uniformly) distributed in the valley, then its volume there would be (400 m)(400 m)d. Requiring these two volumes to be equal, we can solve for d. Thus, d = 25 m. The volume of a small part of the mud over a patch of area of 4.0 m2 is (4.0)d = 100 m3. Since each cubic meter corresponds to a mass of  (1 10 kg m )(5700 m )    3 3 3 6 , 5.70 10 kg V  6 10 kg 10 s M    158 kg s. R 4 V      3 0.158 m /s 42 gal/s. R 4   3    6 50 10 m      3     r         9 1.36 10 kg. m V 3 3   1.91 10 .  8 N  8 2 m        8 9 1.36 10 kg 0.260 kg.

11. 11 1900 kg (stated in the problem), then the mass of that small part of the mud is 5 1.9 10 kg  . 26. (a) The volume of the cloud is (3000 m)(1000 m)2 = 9.4109 m3. Since each cubic meter of the cloud contains from 50 106 to 500 106 water drops, then we conclude that the entire cloud contains from 4.71018 to 4.71019 drops. Since the volume of each drop is 4  to 2 10  m3. (b) Using the fact that 1 L 1 10 cm 1 10 m     part (a) would fill from 2 10  to 2 10  bottles. (c) At 1000 kg for every cubic meter, the mass of water is from The coincidence in numbers between the results of parts (b) and (c) of this problem is due to the fact that each liter has a mass of one kilogram when water is at its normal density (under standard conditions). 27. We introduce the notion of density, m V   and 100 cm = 1 m. (a) The density  of a sample of iron is  7.87 g cm 1000 g    If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and V is the volume of an atom, then 26 9.27 10 7.87 10 kg m   (b) We set V = 4R3/3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find  3 1.18 10 3 4 4       The center-to-center distance between atoms is twice the radius, or 2.82  1010 m. 3 (10106 m)3 = 4.21015 m3, then the total volume of water in a cloud 3 2 10 4 is from  3 3 3 3 , the amount of water estimated in 6 7 2 10  2 10  6 7 to kg. , and convert to SI units: 1000 g = 1 kg, / 3          1 kg 100 cm 1 m     3 3 7870 kg/m .   kg M      29 3 1.18 10 m . V 3 3  1 3         29 3 m 1 3     V      10 1.41 10 m. R

12. smtb98@gmail.com 12 CHAPTER 1 smtb98@gmail.com 28. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom (in the cat) as 10 u  21026 kg, then there are roughly (10 kg)/( 21026 kg)  5 1026 atoms. This is close to being a factor of a thousand greater than Avogadro’s number. Thus this is roughly a kilomole of atoms. 29. The mass in kilograms is 100 16 . piculs 1picul 1gin b which yields 1.747  106 g or roughly 1.75 103 kg. 30. To solve the problem, we note that the first derivative of the function with respect to time gives the rate. Setting the rate to zero gives the time at which an extreme value of the variable mass occurs; here that extreme value is a maximum. (a) Differentiating ( ) 5.00 3.00 20.00 m t t t    dm t dt The water mass is the greatest when / dm dt  (b) At 4.21s, t  the water mass is 0.8 ( 4.21s) 5.00(4.21) m t    (c) The rate of mass change at 2.00s t  is dm dt    (d) Similarly, the rate of mass change at 5.00s t  dm dt     31. The mass density of the candy is gFHG IKJFHG IKJFHG IKJFHG IKJFHG IKJ gin tahil 10 1tahil chee 10 1 chee hoon 03779 . 1hoon g 289 0.8 with respect to t gives 0.2    4.00 3.00. t   1/0.2 or at 0, (4.00/3.00) 4.21s.   3.00(4.21) 20.00 23.2g. g 1kg 60s   3.00 g/s   0.2        4.00(2.00) 0.48g/s 0.48s 1000g 1 min 2.00s t  2 2.89 10 kg/min. is g 1kg 60s   3.00 g/s   0.2    0.101g/s       4.00(5.00) 0.101s 1000g 1 min 2.00s t  3 6.05 10 kg/min.

13. 13 0.0200g 50.0 mm m V          4 3 4 3 4.00 10 g/mm 4.00 10 kg/cm . 3 If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the container when filled to height h is (14.0 cm)(17.0 cm) 238 cm A  is the base area of the container that remains unchanged. Thus, the rate of mass change is given by ( ) (4.00 10 A dt dt dt   32. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20  12 = 240 m2) in addition to a rectangular box (height h´ = 6.0 m and same base). Therefore, 1 2 2  (a) Each dimension is reduced by a factor of 1/12, and we find c h   where , M Ah 2  dM d Ah dh       4 3 2 kg/cm )(238 cm )(0.250 cm/s) 0.0238kg/s 1.43kg/min.      h h A        3 1800 m . V hA h A FHGIKJ  12 3 1  3 3 . 1800 m 10 . m Vdoll (b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore, FHGIKJ  144 c h 33. THINK In this problem we are asked to differentiate between three types of tons: displacement ton, freight ton and register ton, all of which are units of volume. EXPRESS The three different tons are defined in terms of barrel bulk, with 3 1 barrel bulk 0.1415 m 4.0155 U.S. bushels   Thus, in terms of U.S. bushels, we have 4.0155 U.S. bushels 1 displacement ton (7 barrels bulk) 1 barrel bulk  4.0155 U.S. bushels 1 freight ton (8 barrels bulk) 1 barrel bulk       3 1    3 4 3 . 1800 m 6.0 10 m Vminiature  3 (using ). 1 m 28.378 U.S. bushels            28.108 U.S. bushels      32.124 U.S. bushels    4.0155 U.S. bushels 1 barrel bulk  1 register ton (20 barrels bulk) 80.31 U.S. bushels

14. smtb98@gmail.com 14 CHAPTER 1 smtb98@gmail.com ANALYZE (a) The difference between 73 “freight” tons and 73 “displacement” tons is 73(freight tons displacement tons) V       (b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is 73(register tons displacement tons) V      73(32.124 U.S. bushels 28.108 U.S. bushels) 293.168 U.S. bushels 293 U.S. bushels 73(80.31 U.S. bushels 28.108 U.S. bushels)    3.81 10 U.S. bushels  3 3810.746 U.S. bushels LEARN With 1 register ton 1 freight ton 1displacement ton,  found in (b) to be greater than that in (a). This is indeed the case. 34. The customer expects a volume V1 = 20  7056 in3 and receives V2 = 20  5826 in.3, the difference being 1 2 24600 in. V V V      24600 in. 1inch  where Appendix D has been used. 35. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally: 2 peck 11 tuffets = 11 tuffets 22 pecks 1 tuffet   0.50 Imperial bushel 11 tuffets = 11 tuffets 1 tuffet  36.3687 L 11 tuffets = 5.5 Imperial bushel 1 Imperial bushel  36. Table 7 can be completed as follows: (a) It should be clear that the first column (under “wey”) is the reciprocal of the first row – so that 9 and 1 gill = 8.32  106 wey are the last two entries in the first column. (b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 chaldron (that is, the entries along the “diagonal”in the table must be 1’s). To find out how many  we expect the difference 3 , or 3           2.54cm 1L     3 403L V 3 1000 cm        (a) .        5.5 Imperial bushels (b) .        200 L (c) . 10 = 0.900,3 40 = 7.50  102, and so forth. Thus, 1 pottle = 1.56  103 wey

15. 15 chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 1 chaldron = 1 bag. Thus, the next entry in that second column is 1 Similarly, 1 pottle = 1.74  103 chaldron and 1 gill = 9.24  106 chaldron. (c) In the third column (under “bag”), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1 pottle = 2.08  102 bag, and 1 gill = 1.11  104 bag. (d) In the fourth column (under “pottle”), we find 1 chaldron = 576 pottle, 1 bag = 48 pottle, 1 pottle = 1 pottle, and 1 gill = 5.32  103 pottle. (e) In the last column (under “gill”), we obtain 1 chaldron = 1.08  105 gill, 1 bag = 9.02  103 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill. (f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since each bag is 0.1091 m3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m3. 37. The volume of one unit is 1 cm3 = 1106 m3, so the volume of a mole of them is 6.021023 cm3 = 6.021017 m3. The cube root of this number gives the edge length: 5 3 8.4 10 m  . This is equivalent to roughly 8  102 km. 38. (a) Using the fact that the area A of a rectangle is (width) length), we find     40 perch 3.00 acre 1acre   We multiply this by the perch2 rood conversion factor (1 rood/40 perch2) to obtain the answer: Atotal = 14.5 roods. (b) We convert our intermediate result in part (a):  total 580 perch 1perch  Now, we use the feet  meters conversion given in Appendix D to obtain  total 1.58 10 ft 3.281ft  12 12 = 8.33  102.     3.00acre 25.0perch 4.00perch A total        4 perch     2 100perch 2 580 perch . 2      16.5ft     2 5 2 1.58 10 ft . A 2      1m      5 2 4 2 1.47 10 m . A

16. smtb98@gmail.com 16 CHAPTER 1 39. THINK This problem compares the U.K. gallon with U.S. gallon, two non-SI units for volume. The interpretation of the type of gallons, whether U.K. or U.S., affects the amount of gasoline one calculates for traveling a given distance. EXPRESS If the fuel consumption rate is R (in miles/gallon), then the amount of gasoline (in gallons) needed for a trip of distance d (in miles) would be smtb98@gmail.com (miles) d  (gallon) V (miles/gallon) R Since the car was manufactured in U.K., the fuel consumption rate is calibrated based on U.K. gallon, and the correct interpretation should be “40 miles per U.K. gallon.” In U.K., one would think of gallon as U.K. gallon; however, in the U.S., the word “gallon” would naturally be interpreted as U.S. gallon. Note also that since 1 U.K. gallon and 1 U.S. gallon 3.7854118 L  , the relationship between the two is 1 U.S. gallon 1 U.K. gallon (4.5460900 L) 3.7854118 L  ANALYZE (a) The amount of gasoline actually required is 750 miles 18.75 U.K. gallons 18.8 U.K. gallons 40 miles/U.K. gallon This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons. (b) Using the conversion factor found above, this is equivalent to 1.20095 U.S. gallons 18.75 U.K. gallons 1U.K. gallon  LEARN One U.K. gallon is greater than one U.S gallon by roughly a factor of 1.2 in volume. Therefore, 40 mi/U.K. gallon is less fuel-efficient than 40 mi/U.S. gallon. 40. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to simply taking the reciprocal of the number given in Eq.1-9 and rounding off appropriately. Thus, the answer is 6.0  1026. 41. THINK This problem involves converting cord, a non-SI unit for volume, to SI unit. EXPRESS Using the (exact) conversion 1 in. = 2.54 cm = 0.0254 m for length, we have  4.5460900 L        1.20095 U.S. gallons V               22.5 U.S. gallons V

17. 17       0.0254 m 1in     . 1 ft 12in (12in.) 0.3048 m   3 3 3 Thus, Appendix D). ANALYZE The volume of a cord of wood is the conversion factor found above, we obtain for volume (these results also can be found in 1 ft (0.3048 m) 0.0283 m V  (8 ft) (4 ft) (4 ft) 128 ft    3 . Using       3 0.0283 m 1 ft  1cord 128 ft     3 3 3 (128 ft ) 3.625 m V 3       1    3 which implies that . 1 m cord 0.276cord 0.3cord 3.625 LEARN The unwanted units ft3 all cancel out, as they should. In conversions, units obey the same algebraic rules as variables and numbers. 42. (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eq. 1-9, we find 1.6605402 18u = 18u 1u  (b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules: 21 1.4 10 3.0 10  43. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3106 mg/week)/(604800 s/week) = 3.8 mg/s. 44. The volume of the water that fell is   6 3 1.3 10 m .          27 10 kg      26 3.0 10 kg.     46 5 10 . N  26 2             1000 m 1 km 0.0254 m 1in.          2 2 26 km 2.0 in. 26 km 2.0 in. V      6 2 26 10 m 0.0508 m m V   1 10 kg m .   3 3 We write the mass-per-unit-volume (density) of the water as: The mass of the water that fell is therefore given by m = V:  1 10 kg m m         3 3 6 3 9 1.3 10 m 1.3 10 kg.

18. smtb98@gmail.com 18 CHAPTER 1 smtb98@gmail.com 45. The number of seconds in a year is 3.156  107. This is listed in Appendix D and results from the product (365.25 day/y) (24 h/day) (60 min/h) (60 s/min). (a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year. (b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by 10 10 c 6 u-day,  4 10 10 FHG IKJ h 86400 1 u-sec u-day  4 which may also be expressed as 10 u-day 86 . u-sec. 46. The volume removed in one year is V = (75  h   4 2 7 3 , 10 m ) (26 m) FHG 1000 m 2 10 m IKJ  3 c 1 km which we convert to cubic kilometers: V   7 3 3 . 2 10 m 0020 . km 47. THINK This problem involves expressing the speed of light in astronomical units per minute. EXPRESS We first convert meters to astronomical units (AU), and seconds to minutes, using 1000 m 1 km, 1 AU 1.50 ANALYZE Using the conversion factors above, the speed of light can be rewritten as 8 8 3.0 10 m 1 km 3.0 10 m/s s 1000 m      LEARN When expressed the speed of light c in AU/min, we readily see that it takes about 8.3 (= 1/0.12) minutes for sunlight to reach the Earth (i.e., to travel a distance of 1 AU). 48. Since one atomic mass unit is 1u 1.66 10   mole of atoms is about (1.66 10 g)(6.02 10 ) 1g. m   of one mole of atoms in the common Eastern mole is 8 1 min. 10 km, 60 s                     AU 10 km  60 s min     0.12 AU min. c 8 1.50  24 g (see Appendix D), the mass of one  On the other hand, the mass   24 23

19. 19 75g 7.5 m   10g Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is 10g 1.66 10 6.02 10 A N  49. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain 1 1 1 m (b) Similarly, we find 1 1 1 m (c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus,    3.00 r h    (d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) = 1.19  103 m3. 50. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5.95 km. 51. (a) For the minimum (43 cm) case, 9 cubits converts as follows:  m    g 10 u.  23 23 2 2 2 ken 197 . m 2   388 . . 2 m 3 3 3 ken 197 . m 3   765 . . 3 m  2  2 3 5.50 156 ken .       0.43m 1cubit     9cubits 9cubits 3.9m.       0.53m 1cubit     9cubits 9cubits 4.8m. And for the maximum (53 cm) case we have (b) Similarly, with 0.43 m  430 mm and 0.53 m  530 mm, we find 3.9  103 mm and 4.8  103 mm, respectively. (c) We can convert length and diameter first and then compute the volume, or first compute the volume and then convert. We proceed using the latter approach (where d is diameter and  is length).

20. smtb98@gmail.com 20 CHAPTER 1 smtb98@gmail.com 3        0.43m 1 cubit       2 3 3 3 28 cubit 28 cubit 2.2 m . cylinder, min V d 4 Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max = 4.2 m3. 52. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors:     1barn 11 barn 53. THINK The objective of this problem is to convert the Earth-Sun distance (1 AU) to parsecs and light-years. EXPRESS To relate parsec (pc) to AU, we note that when  is measured in radians, it is equal to the arc length s divided by the radius R. For a very large radius circle and small value of , the arc may be approximated as the straight line-segment of length 1 AU. Thus, 1 arcmin 1 arcsec 1 arcsec 60 arcsec 60 arcmin   Therefore, one parsec is 1 AU 1 pc 4.85 10   Next, we relate AU to light-year (ly). Since a year is about 3.16  107 s,  1ly 186,000mi s 3.16  ANALYZE (a) Since 1 pc 2.06 10 AU   , inverting the relation gives 1 pc 1 AU 1 AU 2.06 10 AU   (b) Given that 1AU 92.9 10 mi and 1 ly together lead to   2 m  2 100 hide 1wp 110acre 1hide 4047 m 1acre 25 wp 1 10 .   36    28 1 10                1 2 radian 360        4.85 10 rad  6 .  s     5 . 2.06 10 AU  6       7 12 . 10 s 5.9 10 mi 5            6 4.9 10 pc. 5   12 6 , the two expressions 5.9 10 mi       1ly 10    92.9 10 mi   (92.9 10 mi)    6 6 5 1AU 1.57 10 ly . 12 5.9 mi

21. 21  LEARN Our results can be further combined to give 1pc expression, we readily see that it takes travel a distance of 1 AU to reach the Earth. 54. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3 = 1.639  102 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460 ft2/gal becomes 460ft 1m 460 ft /gal gal 3.28ft    (b) Also, since 1 m3 is equivalent to 1000 L, our result from part (a) becomes 2 2 11.3m 1000L 11.3m /L L 1 m    (c) The inverse of the original quantity is (460 ft2/gal)1 = 2.17  103 gal/ft2. (d) The answer in (c) represents the volume of the paint (in gallons) needed to cover a square foot of area. From this, we could also figure the paint thickness [it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part (b)]. 55. (a) The receptacle is a volume of (40 cm)(40 cm)(30 cm) = 48000 cm3 = 48 L = (48)(16)/11.356 = 67.63 standard bottles, which is a little more than 3 nebuchadnezzars (the largest bottle indicated). The remainder, 7.63 standard bottles, is just a little less than 1 methuselah. Thus, the answer to part (a) is 3 nebuchadnezzars and 1 methuselah. (b) Since 1 methuselah.= 8 standard bottles, then the extra amount is 8  7.63 = 0.37 standard bottle. (c) Using the conversion factor 16 standard bottles = 11.356 L, we have From the above 3.2 ly. 5 y, or about 8.3 min, for Sunlight to 1.57 10 2              2 1gal 3.79L   2 2 11.3 m L.            4 1 1.13 10 m . 3       11.356 L   0.37 standard bottle (0.37 standard bottle) 0.26 L. 16 standard bottles 56. The mass of the pig is 3.108 slugs, or (3.108)(14.59) = 45.346 kg. Referring now to the corn, a U.S. bushel is 35.238 liters. Thus, a value of 1 for the corn-hog ratio would be equivalent to 35.238/45.346 = 0.7766 in the indicated metric units. Therefore, a value of 5.7 for the ratio corresponds to 5.7(0.777)  4.4 in the indicated metric units. 57. Two jalapeño peppers have spiciness = 8000 SHU, and this amount multiplied by 400 (the number of people) is 3.2106 SHU, which is roughly ten times the SHU value for a

22. smtb98@gmail.com 22 CHAPTER 1 smtb98@gmail.com single habanero pepper. More precisely, 10.7 habanero peppers will provide that total required SHU value. 58. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where x = 0.05 m is the increase in horizontal depth per step) 4.57 0.19  However, we can approach this more carefully by noting that if there are N = 4.57/.19  24 rises then under normal circumstances we would expect N 1 = 23 runs (horizontal pieces) in that staircase. This would yield (23)(0.05 m) = 1.15 m, which - to two significant figures - agrees with our first result. 59. The volume of the filled container is 24000 cm3 = 24 liters, which (using the conversion given in the problem) is equivalent to 50.7 pints (U.S). The expected number is therefore in the range from 1317 to 1927 Atlantic oysters. Instead, the number received is in the range from 406 to 609 Pacific oysters. This represents a shortage in the range of roughly 700 to 1500 oysters (the answer to the problem). Note that the minimum value in our answer corresponds to the minimum Atlantic minus the maximum Pacific, and the maximum value corresponds to the maximum Atlantic minus the minimum Pacific. 60. (a) We reduce the stock amount to British teaspoons: 1 breakfastcup = 2 8 1 teacup = 8 2 6 tablespoons = 1 dessertspoon = 2 teaspoons which totals to 122 British teaspoons, or 122 U.S. teaspoons since liquid measure is being used. Now with one U.S cup equal to 48 teaspoons, upon dividing 122/48  2.54, we find this amount corresponds to 2.5 U.S. cups plus a remainder of precisely 2 teaspoons. In other words, 122 U.S. teaspoons = 2.5 U.S. cups + 2 U.S. teaspoons. (b) For the nettle tops, one-half quart is still one-half quart. (c) For the rice, one British tablespoon is 4 British teaspoons which (since dry-goods measure is being used) corresponds to 2 U.S. teaspoons. (d) A British saltspoon is measure is again being used) to 1 U.S. teaspoon.            0.05 m 1.2 m. x N x steps        2 2 = 32 teaspoons teaspoons  2 = 64 teaspoons 6 2 2 24 1 2 British teaspoon which corresponds (since dry-goods